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\(\dfrac{1-x}{3}=\dfrac{2y-1}{8}\)
=>8(1-x)=3(2y-1)
=>8-8x=6y-3
=>-8x-6y=-11
=>8x+6y=11
mà 2x+y=6
nên ta có hệ phương trình:
\(\left\{{}\begin{matrix}8x+6y=11\\2x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8x+6y=11\\8x+4y=24\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2y=-13\\2x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{13}{2}\\2x=6-y=6+\dfrac{13}{2}=\dfrac{25}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{25}{4}\\y=-\dfrac{13}{2}\end{matrix}\right.\)
\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8-6^8.20}\)
\(A=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8-\left(2.3\right)^8.2^2.5}\)
\(A=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8-2^{10}.3^8.5}\)
\(A=\frac{2^{10}.\left(3^8-3^9\right)}{2^{10}.3^8.\left(1-5\right)}=\frac{3^8-3^9}{3^8.\left(-4\right)}=\frac{3^8.\left(1-3\right)}{3^8.\left(-4\right)}=\frac{-2}{-4}=\frac{1}{2}\)
Vậy A = \(\frac{1}{2}\)
\(B=\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(B=\frac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)
\(B=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)
\(B=\frac{2^{19}.3^9+3^9.2^{18}.5}{2^{19}.3^9+2^{20}.3^{10}}\)
\(B=\frac{2^{18}.3^9.\left(2+5\right)}{2^{19}.3^9\left(1+2.3\right)}=\frac{7}{2.7}=\frac{1}{2}\)
Vậy B = \(\frac{1}{2}\)